Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 4 - Dynamics: Newton's Laws of Motion - Problems - Page 102: 23

Answer

The tension in the rope is 1410 N.

Work Step by Step

Arlene's weight is supported by the vertical component of the tension in the left part of the rope plus the vertical component of the tension in the right part of the rope. Let $F_T$ be the tension in each side of the rope. The vertical component is $F_T \times sin(\theta)$. $2F_T \times sin(\theta) = mg$ $F_T = \frac{mg}{2sin(\theta)}$ $F_T = \frac{(50.0 ~kg)(9.80 ~m/s^2)}{2~sin(10.0^{\circ})}$ $F_T = 1410~N$ The tension in the rope is 1410 N.
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